Domain of the function f(x) = frac{x - 3}{(x | vedclass

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Question

(B) For the function $f(x) = \frac{x - 3}{(x - 1)\sqrt{x^2 - 4}}$ to be defined,the expression inside the square root must be strictly greater than ze

Vedclass · Accepted Answer

$( - \infty, - 2) \cup (2, \infty)$

Vedclass · Answer

(B) For the function $f(x) = \frac{x - 3}{(x - 1)\sqrt{x^2 - 4}}$ to be defined,the expression inside the square root must be strictly greater than zero,and the denominator must not be zero.$1$. The condition for the square root is $x^2 - 4 > 0$,which implies $x^2 > 4$. This gives $x > 2$ or $x $2$. The condition for the denominator is $(x - 1)\sqrt{x^2 - 4} 
eq 0$. This implies $x - 1 
eq 0$ (so $x 
eq 1$) and $\sqrt{x^2 - 4} 
eq 0$ (so $x^2 - 4 
eq 0$,which means $x 
eq \pm 2$).$3$. Combining these conditions,we need $x \in ( - \infty, - 2) \cup (2, \infty)$ and $x 
eq 1$. Since $1$ is not in the interval $( - \infty, - 2) \cup (2, \infty)$,the domain is $( - \infty, - 2) \cup (2, \infty)$.

Domain of the function $f(x) = \frac{x - 3}{(x - 1)\sqrt{x^2 - 4}}$ is

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